A vessel contains 1 mole of `O_2` gas (relative molar mass 32) at a temperature T. The pressure of the gas is P. An identical vessel containing one mole of He gas (relative molar mass 4) at temperature 2T has a pressure of

Options

A. `p//8`
B. `p`
C. `2p`
D. `8p`

Answer

To solve the problem, we will use the Ideal Gas Law, which is given by the equation:

\[ PV = nRT \]

where:
- \( P \) = pressure,
- \( V \) = volume,
- \( n \) = number of moles,
- \( R \) = universal gas constant,
- \( T \) = temperature in Kelvin.

### Step-by-Step Solution:

1. **Identify the Variables for the First Vessel (Oxygen)**:
- Number of moles (\( n_1 \)) = 1 mole (given)
- Molar mass of \( O_2 \) = 32 (not directly needed for this calculation)
- Temperature (\( T_1 \)) = \( T \) (given)
- Pressure (\( P_1 \)) = \( P \) (given)
- Volume (\( V \)) = constant (since the vessels are identical)

Using the Ideal Gas Law for the first vessel:
\[ P_1 V = n_1 R T_1 \]
Substituting the known values:
\[ P V = 1 \cdot R \cdot T \]

2. **Identify the Variables for the Second Vessel (Helium)**:
- Number of moles (\( n_2 \)) = 1 mole (given)
- Molar mass of \( He \) = 4 (not directly needed for this calculation)
- Temperature (\( T_2 \)) = \( 2T \) (given)
- Pressure (\( P_2 \)) = ? (we need to find this)

Using the Ideal Gas Law for the second vessel:
\[ P_2 V = n_2 R T_2 \]
Substituting the known values:
\[ P_2 V = 1 \cdot R \cdot (2T) \]

3. **Set the Volumes Equal**:
Since both vessels are identical, the volumes are equal:
\[ P_1 V = P_2 V \]
This means we can set the equations equal to each other:
\[ P V = P_2 V \]

4. **Cancel the Volume**:
Since \( V \) is common in both equations, we can cancel it out:
\[ P = P_2 \]

5. **Relate the Pressures**:
From the second vessel's equation:
\[ P_2 = \frac{R \cdot (2T)}{V} \]
From the first vessel's equation:
\[ P = \frac{R \cdot T}{V} \]
Now, we can express \( P_2 \) in terms of \( P \):
\[ P_2 = 2P \]

### Final Result:
The pressure of the helium gas in the second vessel is:
\[ P_2 = 2P \]